Problem 56
Question
The reaction of \(\mathrm{CO}_{2}\) with sodium phenoxide at \(400 \mathrm{~K}\) results in the formation of (a) salicyaldehdye (b) sodium benzoate (c) benzoic acid (d) sodium salicylate
Step-by-Step Solution
Verified Answer
The reaction produces (d) sodium salicylate.
1Step 1: Understanding the Reaction Context
The question describes a reaction between carbon dioxide (\(\text{CO}_2\)) and sodium phenoxide at a specified temperature. This is an example of the Kolbe-Schmitt reaction, which is a carboxylation process.
2Step 2: Identifying the Reaction Products
In the Kolbe-Schmitt reaction, the reaction of sodium phenoxide with \(\text{CO}_2\) at high temperatures such as \(400 \text{ K}\) primarily results in the formation of a salicylate compound. Specifically, it yields sodium salicylate.
3Step 3: Comparing with Given Options
The product sodium salicylate corresponds to option (d) from the provided choices: (a) salicyaldehyde, (b) sodium benzoate, (c) benzoic acid, and (d) sodium salicylate.
Key Concepts
Sodium Phenoxide ReactionsCarboxylation ProcessSodium Salicylate Production
Sodium Phenoxide Reactions
Sodium phenoxide is an organometallic compound, specifically a type of phenolic salt typically derived from phenol and sodium hydroxide. This compound is crucial in reactions involving nucleophilic substitution and electrophilic aromatic substitutions. In the Kolbe-Schmitt reaction, sodium phenoxide acts as a substrate that reacts with carbon dioxide.
The significance of sodium phenoxide in these reactions is largely due to its ability to act as a nucleophile. This means that it can donate a pair of electrons to an electron-deficient carbon atom, such as the one in carbon dioxide.
The significance of sodium phenoxide in these reactions is largely due to its ability to act as a nucleophile. This means that it can donate a pair of electrons to an electron-deficient carbon atom, such as the one in carbon dioxide.
- During the reaction with \( ext{CO}_2\), sodium phenoxide plays a vital role as the base substrate.
- It is transformed by the reaction's conditions into sodium salicylate, as part of the carboxylation process.
Carboxylation Process
The carboxylation process in the context of the Kolbe-Schmitt reaction involves introducing a carboxyl group \(-COOH\) into an aromatic compound. This process is critical because it enables the formation of aromatic carboxylic acids, which are vital in various industrial and pharmaceutical applications.
In the Kolbe-Schmitt reaction, the interaction BETWEEN carbon dioxide and sodium phenoxide initiates carboxylation. The process requires high temperatures, typically around 400 K, to proceed effectively.
In the Kolbe-Schmitt reaction, the interaction BETWEEN carbon dioxide and sodium phenoxide initiates carboxylation. The process requires high temperatures, typically around 400 K, to proceed effectively.
- Carbon dioxide reacts with sodium phenoxide, causing the aromatic ring to undergo carboxylation.
- The carboxyl group is added ortho to the phenoxide group, resulting in the formation of sodium salicylate.
Sodium Salicylate Production
Sodium salicylate is a vital compound frequently used in various medicinal and industrial applications. Produced through the Kolbe-Schmitt reaction, it represents a classic example of an efficient synthesis pathway via carboxylation of sodium phenoxide.
The manufacture of sodium salicylate is significant because it serves as the precursor to salicylic acid, the active ingredient in aspirin.
The manufacture of sodium salicylate is significant because it serves as the precursor to salicylic acid, the active ingredient in aspirin.
- This process highlights how an organometallic salt, like sodium phenoxide, can be transformed into a valuable chemical compound.
- Once formed, sodium salicylate can be further processed to produce other salicylate-based compounds.
Other exercises in this chapter
Problem 54
Ethylene reacts with \(1 \%\) cold alkaline \(\mathrm{KMnO}_{4}\) to form (a) ethylene glycol (b) oxalic acid (c) \(\mathrm{HCHO}\) (d) ethyl alcohol
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Ethylene reacts with \(1 \%\) cold alkaline \(\mathrm{KMnO}_{4}\) to form (a) ethylene glycol (b) oxalic acid (c) HCHO (d) ethyl alcohol
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When diethyl ether is treated with an excess of \(\mathrm{Cl}_{2}\) in the presence of sunlight, the product formed is (a) \(\mathrm{CH}_{3} \mathrm{CHCl}-\math
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